Depth First Search Algorithm to Compute the Smallest String Star

  • 时间：2020-09-09 13:16:32
  • 分类：网络文摘
  • 阅读：110 次

Given the root of a binary tree, each node has a value from 0 to 25 representing the letters ‘a’ to ‘z’: a value of 0 represents ‘a’, a value of 1 represents ‘b’, and so on.

Find the lexicographically smallest string that starts at a leaf of this tree and ends at the root.

(As a reminder, any shorter prefix of a string is lexicographically smaller: for example, “ab” is lexicographically smaller than “aba”. A leaf of a node is a node that has no children.)

Example 1:
Input: [0,1,2,3,4,3,4]
Output: “dba”
binary-tree1 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Example 2:
Input: [25,1,3,1,3,0,2]
Output: “adz”
binary-tree2 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Example 3:
Input: [2,2,1,null,1,0,null,0]
Output: “abc”
binary-tree3 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Note:
The number of nodes in the given tree will be between 1 and 8500.
Each node in the tree will have a value between 0 and 25.

Recursive Depth First Search to Compute the Smallest String Starting From Leaf

We can start from the Root, perform a recursive DFS (Depth First Search) algorithm, and pass the string along down to the leaves. When we reach a leaf node, we construct the final string, and store the smallest one (alphabetically).

All nodes are visisted exactly once. When we finish searching the tree, we have the answer. The time complexity is O(N). In this particular case, the DFS is also the bruteforce algorithm.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        dfs(root, "");
        return ans;
    }
private:
    string ans = "";
    void dfs(TreeNode* root, string cur) {
        if (!root) return;
        if ((root->left == nullptr) && (root->right == nullptr)) {
            cur = (char)(97 + root->val) + cur;
            if ((ans == "") || (cur < ans)) {
               ans = cur; 
            }
            return;
        }
        dfs(root->left, (char)(97 + root->val) + cur);
        dfs(root->right,(char)(97 + root->val) + cur);
    }
};
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        dfs(root, "");
        return ans;
    }
private:
    string ans = "";
    void dfs(TreeNode* root, string cur) {
        if (!root) return;
        if ((root->left == nullptr) && (root->right == nullptr)) {
            cur = (char)(97 + root->val) + cur;
            if ((ans == "") || (cur < ans)) {
               ans = cur; 
            }
            return;
        }
        dfs(root->left, (char)(97 + root->val) + cur);
        dfs(root->right,(char)(97 + root->val) + cur);
    }
};

Depth First Searching Algorithm using Stack

The Recursion uses implicit stacks. The compiler generates the calling stacks. The recursion code usually is more clean and concise. We can also use a stack to manually emulate the recursive calls.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        if (!root) return "";
        stack<pair<TreeNode*, string>> st;
        st.push({root, ""});
        string ans = "";
        while (!st.empty()) {
            auto p = st.top();
            st.pop();
            if (p.first->right) {
                st.push({p.first->right, (char)(97 + p.first->val) + p.second});
            }            
            if (p.first->left) {
                st.push({p.first->left, (char)(97 + p.first->val) + p.second});
            }
            if ((p.first->left == nullptr) && (p.first->right == nullptr)) {
                string v = (char)(97 + p.first->val) + p.second;
                if ((ans == "") || (v < ans)) {
                    ans = v;   
                }                
            }
        }
        return ans;
    }
};
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        if (!root) return "";
        stack<pair<TreeNode*, string>> st;
        st.push({root, ""});
        string ans = "";
        while (!st.empty()) {
            auto p = st.top();
            st.pop();
            if (p.first->right) {
                st.push({p.first->right, (char)(97 + p.first->val) + p.second});
            }            
            if (p.first->left) {
                st.push({p.first->left, (char)(97 + p.first->val) + p.second});
            }
            if ((p.first->left == nullptr) && (p.first->right == nullptr)) {
                string v = (char)(97 + p.first->val) + p.second;
                if ((ans == "") || (v < ans)) {
                    ans = v;   
                }                
            }
        }
        return ans;
    }
};

During the code interviews, if you are asked this question, it is better to go with the Recursion. But you can follow up with the non-recursive/iterative approach.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
西红柿具有较强的防癌功效　 蔡甸海欣教育　 面对食品安全危机，你应有的态度！　 竹笋的营养与食疗保健功能　 膳食平衡健康新概念“伴侣食品”　 老酸奶调查：由酸奶添明胶，营养价值不高　 惩罚性判罚或许可治食品安全问题之根　 食物中的必需氨基酸和非必需氨基酸　 食品营养价值及食品营养价值的评价　 八大类食品对人体的营养价值分析　
评论列表
添加评论